$$ \,\mathrm{f}\left(\, x\, \right) \equiv \left\{\begin{array}{rcrcccl} 1 & \quad\mbox{if}\quad & 0 & < & x & \leq & a \\[1mm] -1 & \quad\mbox{if}\quad & -a & \leq & x & \leq & 0 \\[1mm] 0 & \quad\mbox{if}\quad &&& \left\vert\, x\, \right\vert & > & a \end{array}\right. $$
Using the fourier transform of $\,\mathrm{f}\left(\, x\, \right)$, calculate the following integral: $$ \,\mathrm{g}\left(\, a,b\, \right) = \int_{0}^{\infty}{\cos\left(\, ax\, \right) - 1 \over x}\, \sin\left(\, bx\,\right)\,\mathrm{d}x\,;\qquad a,b > 0 $$ I found that the fourier transform of $\,\mathrm{f}\left(\, x\, \right)$ is: $$ \widehat{\mathrm{f}}\left(\, \omega\, \right) = {1 - \cos\left(\,\omega a\, \right) \over \pi\mathrm{i}\omega} $$
which leads to: $$ \widehat{\mathrm{f}}\left(\, x\, \right) = {1 - \cos\left(\, ax\, \right) \over \pi\mathrm{i}x}\ \to\ \mathrm{g}\left(\, a,b\, \right) =-\pi\mathrm{i}\int_{0}^{\infty}\,\,\widehat{\mathrm{f}}\left(\, x\, \right) \sin\left(\, bx\, \right)\,\mathrm{d}x $$
but I don't know where to go from here. Thought about using the Plancherel theorem but i'm not sure how.
Please help ! :)
$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\,\mathcal{#1}} \newcommand{\mrm}[1]{\,\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mrm{f}\pars{x} & = \int_{-\infty}^{\infty}\hat{\mrm{f}}\pars{k}\expo{-\ic kx} \,{\dd k \over 2\pi} \quad\imp\quad \hat{\mrm{f}}\pars{k} = \int_{-\infty}^{\infty}\mrm{f}\pars{x}\expo{\ic kx}\,\dd x \end{align}
\begin{align} \hat{\mrm{f}}\pars{k} & = -\int_{-a}^{0}\expo{\ic kx}\,\dd x + \int_{0}^{a}\expo{\ic kx}\,\dd x = -\,{1 - \expo{-\ic ka} \over \ic k} + {\expo{\ic ka} - 1\over \ic k} \\[5mm] & = {2 \over \ic k}\,\bracks{\cos\pars{ka} - 1} \\[5mm] \imp\quad & {\cos\pars{ax} - 1 \over x} = -\,\half\,\ic\int_{-\infty}^{\infty}\mrm{f}\pars{t}\expo{\ic xt}\,\dd t \end{align}
\begin{align} \color{#f00}{\mrm{g}\pars{a,b}} & = \int_{0}^{\infty}{\cos\pars{ax} - 1 \over x}\,\sin\pars{bx}\,\dd x= \half\int_{-\infty}^{\infty}{\cos\pars{ax} - 1 \over x}\,\sin\pars{bx}\,\dd x \\[5mm] & = \half\int_{-\infty}^{\infty}\bracks{% -\,\half\,\ic\int_{-\infty}^{\infty}\mrm{f}\pars{t}\expo{\ic xt}\,\dd t} \sin\pars{bx}\,\dd x \\[5mm] & = -\,{1 \over 4}\ic\int_{-\infty}^{\infty}\mrm{f}\pars{t} \int_{-\infty}^{\infty}\expo{\ic tx} \sin\pars{bx}\,\dd x\,\dd t \\[5mm] & = -\,{1 \over 8}\int_{-\infty}^{\infty}\mrm{f}\pars{t} \bracks{\int_{-\infty}^{\infty}\expo{\ic \pars{t + b}x}\,\dd x - \int_{-\infty}^{\infty}\expo{\ic \pars{t - b}x}\,\dd x}\dd t \\[5mm] & = {1 \over 4}\,\pi\int_{-\infty}^{\infty}\mrm{f}\pars{t}\bracks{% \delta\pars{t - b} - \delta\pars{t + b}}\,\dd t = \color{#f00}{{1 \over 4}\,\pi\bracks{\mrm{f}\pars{b} - \mrm{f}\pars{-b}}} \end{align}